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Math Help - integration

  1. #1
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    integration

    Hi all
    I have a problem that I've nearly finished but I'm stuck on the last bit so I'm hoping someone could help out. The question is

    \int\int (y^2 z^2 + z^2 x^2 + x^2 y^2) \: dS
    Evaluated on the region of z^2 = x^2 + y^2 between z=1 and z=2.

    My workings so far are....
    With z=√(x+y), the integrand should become
    y^2 (x^2 + y^2) + x^2 (x^2 + y^2) + x^2 y^2
    which simplifies to (x^2 + y^2)^2 + x^2 y^2

    Now I have
    \sqrt{2} \int \int ( (x^2 + y^2)^2 + x^2 y^2) \: dA

    To evaluate using polar coordinates, I need to determine the limits of integration.
    Since z ranges from 1 to 2 and the equation of the cone is z=r, I have 1 \leq r \leq 2, and 0\leq \theta \leq 2\pi

    By substituting in x=r*cos(θ) and y=r*sin(θ), and multiplying by 'r' (the Jacobian determinant), I get
    \sqrt{2} \int_{0}^{2\pi} \int_{1}^{2}(r^5 + r^3 \cos(\theta)^2 \sin(\theta)^2) \: dr \: d\theta

    i think the best way is to integrate WRT r first:

    \sqrt{2} \int_{0}^{2\pi} \int_{1}^{2}(r^5 + r^3 \cos(\theta)^2 \sin(\theta)^2) \: dr \: d\theta

    (im going to leave out the \sqrt{2} for now)

    \Rightarrow  \int_{0}^{2\pi} [\frac{r^6}{6} + \frac{r^4}{4}]_{1}^{2} \cos(\theta)^2 \sin(\theta)^2) \: d\theta

    \Rightarrow  \int_{0}^{2\pi} [\frac{32}{3} + 4 \cos(\theta)^2 \sin(\theta)^2)] - [\frac{1}{6} + \frac{1}{4} \cos(\theta)^2 \sin(\theta)^2)] \: d\theta

    I know I then simplify this and integrate WRT theta but this is where I'm running into difficulties so any help at all would be so great.

    Thanks
    Last edited by michaela-donnelly; January 27th 2008 at 10:26 AM. Reason: wrong BB code
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by michaela-donnelly View Post
    Hi all
    I have a problem that I've nearly finished but I'm stuck on the last bit so I'm hoping someone could help out. The question is

    \int\int (y^2 z^2 + z^2 x^2 + x^2 y^2) \: dS
    Evaluated on the region of z^2 = x^2 + y^2 between z=1 and z=2.

    My workings so far are....
    With z=√(x+y), the integrand should become
    y^2 (x^2 + y^2) + x^2 (x^2 + y^2) + x^2 y^2
    which simplifies to (x^2 + y^2)^2 + x^2 y^2

    Now I have
    \sqrt{2} \int \int ( (x^2 + y^2)^2 + x^2 y^2) \: dA

    To evaluate using polar coordinates, I need to determine the limits of integration.
    Since z ranges from 1 to 2 and the equation of the cone is z=r, I have 1 \leq r \leq 2, and 0\leq \theta \leq 2\pi

    By substituting in x=r*cos(θ) and y=r*sin(θ), and multiplying by 'r' (the Jacobian determinant), I get
    \sqrt{2} \int_{0}^{2\pi} \int_{1}^{2}(r^5 + r^3 \cos(\theta)^2 \sin(\theta)^2) \: dr \: d\theta

    i think the best way is to integrate WRT r first:

    \sqrt{2} \int_{0}^{2\pi} \int_{1}^{2}(r^5 + r^3 \cos(\theta)^2 \sin(\theta)^2) \: dr \: d\theta

    (im going to leave out the \sqrt{2} for now)

    \Rightarrow  \int_{0}^{2\pi} [\frac{r^6}{6} + \frac{r^4}{4}]_{1}^{2} \cos(\theta)^2 \sin(\theta)^2) \: d\theta

    \Rightarrow  \int_{0}^{2\pi} [\frac{32}{3} + 4 \cos(\theta)^2 \sin(\theta)^2)] - [\frac{1}{6} + \frac{1}{4} \cos(\theta)^2 \sin(\theta)^2)] \: d\theta

    I know I then simplify this and integrate WRT theta but this is where I'm running into difficulties so any help at all would be so great.

    Thanks
    could you fix the LaTeX please? we use [tex] tags here, not [tex]
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    could you fix the LaTeX please? we use [tex] tags here, not [tex]
    I knew it wasn't correct but I'd no idea, until you told me, how to change the BB Code. Thanks for the tip, much appreciated!
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  4. #4
    Global Moderator

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    Quote Originally Posted by michaela-donnelly View Post
    Hi all
    I have a problem that I've nearly finished but I'm stuck on the last bit so I'm hoping someone could help out. The question is

    \int\int (y^2 z^2 + z^2 x^2 + x^2 y^2) \: dS
    Evaluated on the region of z^2 = x^2 + y^2 between z=1 and z=2.
    Let R = \{(r,\theta) \in \mathbb{R}^2 : 1\leq r\leq 2 \mbox{ and }0\leq \theta \leq 2\pi \}. Define \bold{G}:R\mapsto \mathbb{R} by \bold{G}(r,\theta) = (r\cos\theta, r\sin \theta, r). Which means the integral is, \int_0^{2\pi} \int_1^2 f(r\cos \theta, r\sin \theta, r) \left| \partial_r \bold{G} \times \partial_{\theta} \bold{G}\right| ~ dA
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